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This might be a very stupid question, but I can't seem to figure this out and I'm going nuts thinking about it.

Assuming you are trying to adjust the fixed bias of a tetrode push pull pair in a tube amp. In many amps, for convenience, the cathode will have a 1 ohm resistor connecting it to the ground, so you can easily use a multimeter to measure the voltage and thus the current flowing through the tetrode.

My confusion arises when trying to comprehend how the resistance of the multimeter probes themselves would contribute to this measurement.

I realize the probes are likely not thin enough wire or long enough to make any difference, but assume the resistance of each probe wire is also 1 ohm for this scenario.

Would it make any difference here?

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  • \$\begingroup\$ the difference would be the resistance (impedance) of the multimeter \$\endgroup\$ – jsotola Dec 9 '19 at 3:39
  • \$\begingroup\$ When measuring voltage with a meter, effectively no current is drawn, so there resistance of the probes is irrelevant. That said, meters vary in how perfect they are, and there may also be substantial safety issues in probing tube gear which is either live or has charge retained on supply capacitors. \$\endgroup\$ – Chris Stratton Dec 9 '19 at 3:56
  • \$\begingroup\$ How much current is flowing through the probes? It matters. If you are going to work with Ohm's Law, you need to know why it matters. \$\endgroup\$ – Harper - Reinstate Monica Dec 9 '19 at 4:17
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If you were using an analog multimeter with, say, 20,000\$\Omega\$/V on a 0-150mV F.S. range, the meter looks like an approximately 3K resistor.

Thus the voltage across the 1\$\Omega\$ resistor will be a bit less since it is shunted by about 3K (3002 ohms if you count the leads) and the voltage making it to the 3000 ohm meter movement is a bit less as well (3000/3002 of the voltage across the 1\$\Omega\$ resistor) because 1/3002 of that voltage is dropped across each lead.

schematic

simulate this circuit – Schematic created using CircuitLab

In total then, in this example, the voltage read at the meter is lower than ideal by about 0.1%, which is going to be totally insignificant compared to the tolerance of the resistors and the accuracy of the meter.

A typical digital multimeter will have an input resistance of something like 10M\$\Omega\$ on the 199mV scale so the effect will be several thousand times less, so even with a very accurate resistor and meter it's negligible. Not to mention that the leads are probably closer to 0.2\$\Omega\$ than 1\$\Omega\$.

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  • \$\begingroup\$ Thank you!!! :) \$\endgroup\$ – cat pants Dec 24 '19 at 3:21
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if each one is 1ohm they will a difference if the volt meter impedance itself is very low, normally when measuring voltage you want a large resistance on your volt meter(they are usually 10 Megaohm or even larger) if the volt meter impedance is low enough to be comparable to the probes, you already have a useless volt meter

schematic

simulate this circuit – Schematic created using CircuitLab

let's say you got this circuit right here, if you apply kirchoff law's for current at the R1 node there is a current that your circuit provides to this resistor and if you do this with and without the voltmeter the value will be different, this is called "loading" the circuit, if the voltmeter itself will draw too much current from the circuit. Not only the voltage on R1 will change, but all voltages in the circuit will change, so most voltmeters have high resistance to avoid loading the circuit you measure.

now if the voltmeter impedance along with the probe resistance was close to the target resistor you are measuring it would "load" the circuit.

most cases the probes themselves do not make a difference because the voltmeter has a very large input impedance.

edit: changed the voltmeter impedance connection

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    \$\begingroup\$ Good, but draw the voltmeter with the 10 M in parallel with the ideal voltmeter, not series. An ideal voltmeter has infinite impedance so in the circuit as drawn now, no current will flow. \$\endgroup\$ – tomnexus Dec 9 '19 at 5:26
  • \$\begingroup\$ that indeed is better. \$\endgroup\$ – Juan Dec 9 '19 at 5:41
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You are presuming the only resistance of the meter is the probes. Actually, a digital voltmeter would have several megaohms of impedance.

Since the probes are in series with the meter, their resistance is tiny by comparison, and thus, inconsequential.

If you have selected ammeter mode (maybe you did this because the endgame valur you are after is amps), don't do that becase it defeats the purpose of using that 1 ohm resistor as an ammeter shunt! ...but if you did, the meter would be near 0 ohms and so your leads had better have some resistance! However they don't, so don't do this.

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My Electrinics 101 professor once said that you can't measure a circuit's voltage, without affecting the voltage level (very slightly), and then went on to demonstrate that even digital multimeters & their probes contain resistance. The probes have very low resistance (under .5 ohm), and the meter has a very high resistance (usually in excess of 10M ohm.) The point was that usually you can ignore the slight voltage drop caused by the meter, but to be aware that it exists. In my nearly 40 years of professional electronics experience, I've never had to take a meter's voltage drop into account.

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